Calculate the mass of the salt formed by the interaction of acetic acid weighing 15 g with an excess of limestone.

The reaction of calcium carbonate with acetic acid is described by the following chemical reaction equation:

CaCO3 + 2CH3COOH = Ca (CH3COO) 2 + CO2 + H2O;

The reaction of two moles of acid leads to the formation of one mole of salt.

Let’s find the amount of acid.

Its molar mass is:

M CH3COOH = 12 x 2 + 16 x 2 + 4 = 60 grams / mol;

The amount of substance will be:

N CH3COOH = 15/60 = 0.25 mol;

Calculate the mass 0.25 / 2 = 0.125 mol of Ca (CH3COO) 2.

Its molar mass is:

M Ca (CH3COO) 2. = 40 + (12 x 2 + 16 x 2 + 3) x 2 = 158 grams / mol;

The mass of salt will be:

m Ca (CH3COO) 2. = 0.125 x 158 = 19.75 grams;